(A) The Arrhenius equation is given by $k = A \ e^{-E_{a} / RT}$.Given that $E_{a} = 2.303 \ RT \ J \ mol^{-1}$.Substituting the value of $E_{a}$ in t

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(A) The Arrhenius equation is given by $k = A \ e^{-E_{a} / RT}$.Given that $E_{a} = 2.303 \ RT \ J \ mol^{-1}$.Substituting the value of $E_{a}$ in the equation:$k = A \ e^{-(2.303 \ RT) / RT}$$k = A \ e^{-2.303}$Since $e^{-2.303} = 10^{-1}$,we have:$k = A \times 10^{-1}$Therefore,the ratio of the rate constant to the Arrhenius factor is $\frac{k}{A} = 10^{-1}$.

Class 12 Chemistry Chemical Kinetics Collision theory, Energy of activation and Arrhenius equation

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The activation energy for a reaction at the temperature $T \ K$ was found to be $2.303 \ RT \ J \ mol^{-1}$. The ratio of the rate constant to the Arrhenius factor is

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$10^{-1}$
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$10^{-2}$
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$2 \times 10^{-3}$
D
$2 \times 10^{-2}$

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Chemical Kinetics

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Collision theory, Energy of activation and Arrhenius equation

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The rate constant $(k)$ of a reaction is measured at different temperatures $(T),$ and the data are plotted in the given figure. The activation energy of the reaction in $kJ\, mol^{-1}$ is :
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MediumJEE MAIN 2020

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The activation energy for a reaction at the temperature $T \ K$ was found to be $2.303 \ RT \ J \ mol^{-1}$. The ratio of the rate constant to the Arrhenius factor is

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If the definition of the temperature coefficient of the reaction holds good for a reaction between $27^{\circ} C$ and $37^{\circ} C$,the activation energy for the reaction in $kJ \cdot mol^{-1}$ is

When the temperature changes from $293 \ K$ to $313 \ K$,the rate of a certain reaction becomes four times. Find the activation energy of the reaction in $kJ \ mol^{-1}$. $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$

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Fill in the blanks:
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$(b)$ Boltzmann and Maxwell used ......... to explain the rate of reaction.

The rate constant $(k)$ of a reaction is measured at different temperatures $(T),$ and the data are plotted in the given figure. The activation energy of the reaction in $kJ\, mol^{-1}$ is :
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